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Formula
Solve the equation
Answer
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Graph
$y = \dfrac { 1 } { 2 } + \dfrac { 1 } { 5 }$
$y = \dfrac { 1 } { x }$
Asymptote
$y = 0$, $x = 0$
Standard form
$y = \dfrac { 1 } { x }$
Domain
$y \neq 0$
Range
$x \neq 0$
$\dfrac{ 1 }{ 2 } + \dfrac{ 1 }{ 5 } = \dfrac{ 1 }{ x }$
$x = \dfrac { 10 } { 7 }$
Solve the fractional equation
$\color{#FF6800}{ \dfrac { 1 } { 2 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 } { 5 } } = \color{#FF6800}{ \dfrac { 1 } { x } }$
$ $ Reverse the left and right terms of the equation (or inequality) $ $
$\color{#FF6800}{ \dfrac { 1 } { x } } = \color{#FF6800}{ \dfrac { 1 } { 2 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 } { 5 } }$
$\color{#FF6800}{ \dfrac { 1 } { x } } = \color{#FF6800}{ \dfrac { 1 } { 2 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 } { 5 } }$
$ $ If $ \frac{a(x)}{b(x)} = c(x) $ is valid, it is $ \begin{cases} a(x) = b(x) c(x) \\ b(x) \ne 0 \end{cases}$
$\begin{cases} \color{#FF6800}{ 1 } = \color{#FF6800}{ x } \left ( \color{#FF6800}{ \dfrac { 1 } { 2 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 } { 5 } } \right ) \\ \color{#FF6800}{ x } \neq \color{#FF6800}{ 0 } \end{cases}$
$\begin{cases} \color{#FF6800}{ 1 } = \color{#FF6800}{ x } \left ( \color{#FF6800}{ \dfrac { 1 } { 2 } } \color{#FF6800}{ + } \color{#FF6800}{ \dfrac { 1 } { 5 } } \right ) \\ \color{#FF6800}{ x } \neq \color{#FF6800}{ 0 } \end{cases}$
$ $ Simplify the expression $ $
$\begin{cases} \color{#FF6800}{ 1 } = \color{#FF6800}{ \dfrac { 7 x } { 10 } } \\ \color{#FF6800}{ x } \neq \color{#FF6800}{ 0 } \end{cases}$
$\begin{cases} \color{#FF6800}{ 1 } = \color{#FF6800}{ \dfrac { 7 x } { 10 } } \\ x \neq 0 \end{cases}$
$ $ Reverse the left and right terms of the equation (or inequality) $ $
$\begin{cases} \color{#FF6800}{ \dfrac { 7 x } { 10 } } = \color{#FF6800}{ 1 } \\ x \neq 0 \end{cases}$
$\begin{cases} \color{#FF6800}{ \dfrac { 7 x } { 10 } } = \color{#FF6800}{ 1 } \\ x \neq 0 \end{cases}$
$ $ If $ \frac{a(x)}{b(x)} = c(x) $ is valid, it is $ \begin{cases} a(x) = b(x) c(x) \\ b(x) \ne 0 \end{cases}$
$\begin{cases} \begin{cases} \color{#FF6800}{ 7 } \color{#FF6800}{ x } = \color{#FF6800}{ 10 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \\ \color{#FF6800}{ 10 } \neq \color{#FF6800}{ 0 } \end{cases} \\ x \neq 0 \end{cases}$
$\begin{cases} \begin{cases} \color{#FF6800}{ 7 } \color{#FF6800}{ x } = \color{#FF6800}{ 10 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \\ \color{#FF6800}{ 10 } \neq \color{#FF6800}{ 0 } \end{cases} \\ \color{#FF6800}{ x } \neq \color{#FF6800}{ 0 } \end{cases}$
$ $ If there is a system of equations (inequality) in the system of equations (inequality), take it out. $ $
$\begin{cases} \color{#FF6800}{ 7 } \color{#FF6800}{ x } = \color{#FF6800}{ 10 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \\ \color{#FF6800}{ 10 } \neq \color{#FF6800}{ 0 } \\ \color{#FF6800}{ x } \neq \color{#FF6800}{ 0 } \end{cases}$
$\begin{cases} \color{#FF6800}{ 7 } \color{#FF6800}{ x } = \color{#FF6800}{ 10 } \color{#FF6800}{ \times } \color{#FF6800}{ 1 } \\ \color{#FF6800}{ 10 } \neq \color{#FF6800}{ 0 } \\ \color{#FF6800}{ x } \neq \color{#FF6800}{ 0 } \end{cases}$
$ $ Simplify the expression $ $
$\begin{cases} \color{#FF6800}{ 7 } \color{#FF6800}{ x } = \color{#FF6800}{ 10 } \\ \color{#FF6800}{ 10 } \neq \color{#FF6800}{ 0 } \\ \color{#FF6800}{ x } \neq \color{#FF6800}{ 0 } \end{cases}$
$\begin{cases} \color{#FF6800}{ 7 } \color{#FF6800}{ x } = \color{#FF6800}{ 10 } \\ 10 \neq 0 \\ x \neq 0 \end{cases}$
$ $ Solve a solution to $ x$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 10 } { 7 } } \\ 10 \neq 0 \\ x \neq 0 \end{cases}$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 10 } { 7 } } \\ \color{#FF6800}{ 10 } \neq \color{#FF6800}{ 0 } \\ \color{#FF6800}{ x } \neq \color{#FF6800}{ 0 } \end{cases}$
$ $ Substitute $ x = \dfrac { 10 } { 7 } $ for unresolved equations or inequalities $ $
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 10 } { 7 } } \\ \color{#FF6800}{ 10 } \neq \color{#FF6800}{ 0 } \\ \color{#FF6800}{ \dfrac { 10 } { 7 } } \neq \color{#FF6800}{ 0 } \end{cases}$
$\begin{cases} x = \dfrac { 10 } { 7 } \\ \color{#FF6800}{ 10 } \neq \color{#FF6800}{ 0 } \\ \dfrac { 10 } { 7 } \neq 0 \end{cases}$
$ $ There are infinitely many solutions if both sides of $ \ne $ are different. $ $
$\begin{cases} x = \dfrac { 10 } { 7 } \\ \text{There are countless solutions} \\ \dfrac { 10 } { 7 } \neq 0 \end{cases}$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 10 } { 7 } } \\ \text{There are countless solutions} \\ \color{#FF6800}{ \dfrac { 10 } { 7 } } \neq \color{#FF6800}{ 0 } \end{cases}$
$ $ Ignore the cases where the system of equations where there are infinitely many solutions. $ $
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 10 } { 7 } } \\ \color{#FF6800}{ \dfrac { 10 } { 7 } } \neq \color{#FF6800}{ 0 } \end{cases}$
$\begin{cases} x = \dfrac { 10 } { 7 } \\ \color{#FF6800}{ \dfrac { 10 } { 7 } } \neq \color{#FF6800}{ 0 } \end{cases}$
$ $ There are infinitely many solutions if both sides of $ \ne $ are different. $ $
$\begin{cases} x = \dfrac { 10 } { 7 } \\ \text{There are countless solutions} \end{cases}$
$\begin{cases} \color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 10 } { 7 } } \\ \text{There are countless solutions} \end{cases}$
$ $ Ignore the cases where the system of equations where there are infinitely many solutions. $ $
$\color{#FF6800}{ x } = \color{#FF6800}{ \dfrac { 10 } { 7 } }$
$ $ 그래프 보기 $ $
Graph
Solution search results
search-thumbnail-a sdoupIuai apS
$1.$ $2.$ $3.$ 
$\dfrac {1} {x}+\dfrac {x} {2}=\dfrac {11} {6}$ $\dfrac {2} {x}-\dfrac {x+1} {5}=-\dfrac {4} {5}$ $\dfrac {2} {x-3}+\dfrac {x} {2}=-\dfrac {1} {2}$
7th-9th grade
Algebra
search-thumbnail-
$1$ $2.$ $3$ 
$\dfrac {1} {x}+\dfrac {x} {2}=\dfrac {11} {6}$ $\dfrac {2} {x}-\dfrac {x+1} {5}=-\dfrac {4} {5}$ $\dfrac {2} {x-3}+\dfrac {x} {2}=-\dfrac {1} {2}$
7th-9th grade
Algebra
search-thumbnail-Question $4$ 
If $x$ is $6$ what $|s$ 
$\dfrac {1} {3}x$ 
$1frac\left(1\right)\left(3\right)x$
1st-6th grade
Other
search-thumbnail-$11.$ Question $11$ 
Solve the $:$ $folloMlng'$ $0<θ<90^{°}$ 
$\left(1\right)$ $2sin^{2}θ=1\right)$ $\left(rac\left(3\right)\left(2\right)\right)$ 
$\left(11\right)$ $3tan^{2}θ+2=3$ 
$\left(111\right)cos^{2}θ$ $11rac\left(1\right)\left(4\right)\right)=$ 
$c\left(1\right)\left(4\right)\right)=11113c\left(1\right)\left(2\right)\right)$
10th-13th grade
Trigonometry
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